$ g(x) = \int_{0}^{x}(5t^2 - t)\,dt\,$ $ g\,'(2)\, =$
Answer: The Fundamental Theorem of Calculus If $~ g(x)=\int_a^xf(t)\,dt\,$, then $~g^\prime (x)=f(x)\,$ This only works if $f$ is continuous on $[a,b]$. Thankfully, the function $f(t) = 5t^2 -t$ is continuous on $[0,2]$. Applying the theorem We're given: $ g(x) = \int_{0}^{x}(5t^2 - t)\,dt\,$ So the theorem tells us: $ g\,^\prime(x) =5x^2 - x$ Evaluating $g'(2)$ $ g'(2)= 5(2)^2 - 2 = 18$ The answer: $g'(2)=18$